Manuscript III@2 · III.tex · section · 82dad3e5

Past-trivial gauges

The coupled first-order jet and retarded harmonic section › Past-trivial gauges · lines 1101–1138

III@2#eq:retarded-inverse, III@2#eq:retarded-compensator

Let

G_{R} = E_{+} (T^{*} M) .

For $F \in E_{+}$ , define $R_{□} F$ to be the solution of

□_{η} R_{□} F = F

with zero data on a slice preceding $supp F$ . The forced Cauchy lemma constructs compatible solutions on all finite intervals, and finite propagation gives $R_{□} F \in E_{+}$ . The prescription is independent of the earlier slice.

We have

□_{η} R_{□} F = F, R_{□} □_{η} a = a (a \in E_{+}) .

The second equality follows because the difference is homogeneous with zero past data.

Define

M_{R} = {h \in C^{\infty} (M; S^{2} T^{*} M) : (D h)^{♯} \in E_{+} (T M)},

and

w_{R} (h) = R_{□} (D h)^{♯}, Π_{R} h = h - K w_{R} (h) .

In $K w_{R} (h)$ , lower the index first.

Displayed formulas in this passage

III@2#eq:retarded-inverse · line 1117
$□_{η} R_{□} F = F, R_{□} □_{η} a = a (a \in E_{+}) .$
III@2#eq:retarded-compensator · line 1132
$w_{R} (h) = R_{□} (D h)^{♯}, Π_{R} h = h - K w_{R} (h) .$
unlabeled · line 1104
$G_{R} = E_{+} (T^{*} M) .$
unlabeled · line 1108
$□_{η} R_{□} F = F$
unlabeled · line 1126
$M_{R} = {h \in C^{\infty} (M; S^{2} T^{*} M) : (D h)^{♯} \in E_{+} (T M)},$

Past-trivial gauges

Displayed formulas in this passage

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